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Quest HW 8

# Quest HW 8 - roberts(eer474 Quest HW 8 seckin(56425 This...

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roberts (eer474) – Quest HW 8 – seckin – (56425) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the linearization of f ( x ) = 1 3 + x at x = 0. 1. L ( x ) = 1 3 - 1 3 x 2. L ( x ) = 1 3 parenleftBig 1 - 1 6 x parenrightBig correct 3. L ( x ) = 1 3 parenleftBig 1 - 1 3 x parenrightBig 4. L ( x ) = 1 3 parenleftBig 1 + 1 6 x parenrightBig 5. L ( x ) = 1 3 parenleftBig 1 + 1 6 x parenrightBig 6. L ( x ) = 1 3 + 1 3 x Explanation: The linearization of f is the function L ( x ) = f (0) + f (0) x . But for the function f ( x ) = 1 3 + x = (3 + x ) 1 / 2 , the Chain Rule ensures that f ( x ) = - 1 2 (3 + x ) 3 / 2 . Consequently, f (0) = 1 3 , f (0) = - 1 6 3 , and so L ( x ) = 1 3 parenleftBig 1 - 1 6 x parenrightBig . 002 10.0 points Use linear appproximation to estimate the value of 17 1 / 4 . ( Hint : (16) 1 / 4 = 2.) 1. 17 1 / 4 65 32 correct 2. 17 1 / 4 2 3. 17 1 / 4 31 16 4. 17 1 / 4 33 16 5. 17 1 / 4 63 32 Explanation: Set f ( x ) = x 1 / 4 , so that f (16) = 2 as the hint indicates. Then df dx = 1 4 x 3 / 4 . By differentials, therefore, we see that f ( a + Δ x ) - f ( a ) df dx vextendsingle vextendsingle vextendsingle x = a Δ x = Δ x 4 a 3 / 4 . Thus, with a = 16 and Δ x = 1, 17 1 / 4 - 2 = 1 32 . Consequently, 17 1 / 4 65 32 . 003 10.0 points If f is the function defined on [ - 4 , 4] by f ( x ) = x + | x | - 4 , which of the following properties does f have? A. Absolute minimum at x = 0 . B. Differentiable at x = 0 .

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roberts (eer474) – Quest HW 8 – seckin – (56425) 2 1. neither of them 2. both of them 3. A only correct 4. B only Explanation: Since | x | = braceleftbigg x , x 0 , - x , x < 0 , we see that f ( x ) = x + | x |- 4 = braceleftbigg 2 x - 4 , x 0 , - 4 , x < 0 . Thus on [ - 4 , 4] the graph of f is 2 4 - 2 - 4 2 4 - 2 - 4 Consequently: A. True: by inspection. B. False: f is continuous, but not differen- tiable at x = 0 . 004 10.0 points If f is the function whose graph is given by 2 4 6 2 4 6 which of the following properties does f NOT have? 1. differentiable at x = 2 correct 2. local maximum at x = 4 3. f ( x ) < 0 on ( - 1 , 2) 4. lim x 2 + f ( x ) = lim x 2 - f ( x ) 5. lim x 4 f ( x ) = 4 Explanation: The given graph has a removable disconti- nuity at x = 4.
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Quest HW 8 - roberts(eer474 Quest HW 8 seckin(56425 This...

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